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0

Yeaaahh, I got it! :D Great thanks to David C. Ullrich!!! Counterexample Given the Hilbert space $\mathbb{C}^4$. Regard the matrices: $$N:=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\oplus\begin{pmatrix}0&0\\0&0\end{pmatrix}\quad N':=\begin{pmatrix}0&0\\0&0\end{pmatrix}\oplus\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Then they are ...


3

Thank you for the comments. I was able to use them to provide an answer to When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?. I'll restate the result from there as it provides an example: Consider the measure space $L^2([0,1])$ with uniform lebesgue measure. Define the functions (Haar functions) $f_{2^i + k}$ by ...


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If one of the members of the product was NOT invertible,it would map a sequence of vectors, that doesn't converge to zero, to a sequence converging to zero.But then the product would also do that, making the product non-invertible.


1

With a clear domain definition, and $\|\frac{1}{\Delta t}\{U_{\alpha}(\Delta t)-1_\alpha\}\varphi_{\alpha}-H_{\alpha}\varphi_{\alpha}\|=\|\frac{1}{\Delta t}\int_{0}^{\Delta t}(U_{\alpha}(t)-1_\alpha)H_{\alpha}\varphi_\alpha dt\|$, can you now better establish convergence?


3

Another Banach-algebra structure on the Hilbert space $\mathsf{hs}(H)$ of Hilbert-Schmidt operators on a Hilbert space $H$ is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra.


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Left Shift Denote for shorthand: $$1_0(0,x_1,\ldots):=(0,x_1,\ldots)$$ For the modulus: $$L^*L=RL=1_0\implies|L|=1_0$$ For the argument: $$L=U|L|\implies U=1_0,1,\ldots$$ So it admits one. Right Shift For the modulus: $$R^*R=LR=1\implies|R|=1$$ For the argument: $$R=U|R|\implies U=R$$ So it admits none. Reference For much more details: Polar ...


1

For your second question, the answer is always yes. More generally, given any Banach algebra $A$, let $\tilde{A}=A\oplus \mathbb{C}$ be its unitization (with norm $\|(a,z)\|=\|a\|+|z|$). Given $a\in A$, let $L_a\in B(\tilde{A})$ be left multiplication by $a$. Then $a\mapsto L_a$ is an isometric isomorphism from $A$ to a subalgebra of $B(\tilde{A})$. ...


2

As noted in my comment, $E$ has to be a Hilbert space. To see this, fix a linear functional $\varphi \in E'$ with $\Vert \varphi \Vert = 1$. For $z \in E$, define $$ A_z : E \to E, x\mapsto \varphi(x) \cdot z. $$ It is not hard to see that $E \to B(E), z \mapsto A_z$ is linear and isometric. Thus, if $B(E) = H$ is a Hilbert space, we see that $E$ is ...


4

Let $J $ be an index for the cardinality of an orthonormal basis of $H $. Then $H $ is isometrically isomorphic to $\ell^2 (J) $, so it is enough to discuss the problem on this latter space. Define the product $fg $ pointwise, i.e. $fg (j):=f (j)g (j) $. The question is whether this product stays in $\ell^2$, and whether the norm is submultiplicative. We ...


2

I know a very special case, and I guess it is extensible. Suppose that $H$ is Hilbert space such that $H\cong (H_1\hat{\otimes}(H_2)^*)^*$, where $\hat{\otimes}$ is projective tensor product, and $H_1,H_2$ is some Hilbert spaces. For any Hilbert space $\mathcal{H}$, always $\mathcal{H}^{**}=\mathcal{H}$. Also for two Banach space $E,F$ always ...


1

Let $\mathcal S$ be an orthonormal basis of eigenvectors for $|A|$ (this exists, since $|A|$ is trace-class and thus compact). Let $\mathcal T$ be any other orthonormal basis. For any $\sigma\in\mathcal S$, we have $$ |A|\sigma=s_\sigma(A)\,\sigma $$ and $\sum_\sigma s_\sigma(A)=\sum_\sigma\langle |A|\sigma,\sigma\rangle<\infty$. Then, for any $B\in ...


1

Yes. The functional calculus preserves approximation by polynomials. So, from $$ T^n=\begin{bmatrix}A^n&0\\0&B^n\end{bmatrix}, $$ you get that $$ p(T)=\begin{bmatrix}p(A)&0\\0&p(B)\end{bmatrix} $$ for any polynomial $p$. Now using a sequence of polynomials that converges uniformly to $f$, you get that $$ ...


3

Left/right shift operators are the standard examples, but I personally think that the multiplication operator is the easiest way to see that there may be something else in the spectrum besides eigenvalues. Consider a multiplication operator $A_c$ on $\ell^\infty$ ($c\in\ell^\infty$) $$ (A_c x)_n=c_n x_n. $$ The inverse if exists is clearly a multiplication ...


7

If something is non-invertible, there's two (non-disjoint) possibilities: it fails to be injective, or it fails to be surjective. In finite dimension, these are the same, but in infinite-dimensional spaces, weird things can happen. If it fails to be injective, there's $x \ne y$ such that $(T - \lambda I)(x) = (T - \lambda I)(y)$. So $(T - \lambda I)(x - y) ...


6

For finite-dimensional vector spaces, injectivity and surjectivity are equivalent. That's not the case for an arbitrary Hilbert space. The classic examples are the left- and right-shift operators $L, R:\ell^2 \to \ell^2$, given by \begin{align*} L(x_1, x_2, \dots) &= (x_2, \dots) \\ R(x_1, x_2, \dots) &= (0, x_1, x_2, \dots). \end{align*} The map $L$ ...


3

$T-\lambda I$ being non-invertible does not imply there is a non-zero $x$ with $(T-\lambda I)x=0$. That is true when $H$ is finite-dimensional, but not necessarily when $H$ is infinite-dimensional. The classic counterexample is the right-shift operator $R:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$. Take a look at the Wikipedia article on the notion of ...


1

Although $0 \le A \le B$ implies $\overline{{\cal R}(A)} \subseteq \overline{{\cal R}(B)}$, it's not true without the closures. For a counterexample, take $L^2[0,1]$. Let $A$ be multiplication by $x$ (i.e. $A f(x) = x f(x)$), and $B = A + u u^*$ where $u(t) = t^{1/4}$ and $u^*$ is the corresponding linear functional, i.e. $$ B f(x) = A f(x) + u^*(f) u(x) = ...


1

This is always true, as $\Omega \setminus (\Omega_1\cup \Omega_2)$ can only contain the boundaries of $\Omega_1$ and $\Omega_2$, which have zero measure due to the regularity assumptions on the domains.


0

Weak convergence in $H=H^1_0$ implies strong convergence in $L^2$ because $H^1_0$ is compactly embedded in $L^2$, see: https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem


1

The implication $0\leq A\leq B$ $\implies $ $\mathcal RA\subset\mathcal RB$ can be proven as follows. From $0\leq A\leq B$, we easily see that if $Bx=0$, then $$ 0\leq\langle Ax,x\rangle\leq\langle Bx,x\rangle=0, $$ so $A^{1/2}x=0$, and thus $Ax=0$. In other words, $\ker B\subset\ker A$. Then $$ \overline{\mathcal RA}=(\ker A)^\perp\subset(\ker ...


0

No, no, and no. No, $E$ is not even a Hilbert space! If we say $d\mu = w dt$ then the completion of $E$ is $H=L^2(\mu)$, which is a Hilbert space. No, $H$ is not a reproducing-kernel Hilbert space. It can't be, since it's not even a space of functions! (More formally, the map $f\mapsto f(0)$ is not even defined on $H$.) It still could be that there is a ...


1

Presumably, you mean for $H$ to be a vector space over $\Bbb C$. Note that $\operatorname{Herm}(H^k)$ is a vector space over $\Bbb R$ (as opposed to $\Bbb C$), and that $\dim \operatorname{Herm}(H^k) = k^2$. From there, we can simply apply your earlier reasoning to find that $$ \mathrm{dim}(\mathrm{Sym}(\mathrm{Herm}(H^k)^{\otimes N})) = \binom{N + k^2 - ...


1

If $\alpha\neq 1$, then it fails for $x=y$, since the right hand side is $0$, and the left hand side is positive.


0

Existence For complex sums: $$\sum_\sigma\langle A\sigma,\sigma\rangle\in\mathbb{C}\iff\sum_\sigma|\langle A\sigma,\sigma\rangle|<\infty$$ By polar decomposition: $$|\langle A\sigma,\sigma\rangle|\leq\||A|^{1/2}\sigma\|\cdot\||A|^{1/2}J^*\sigma\|$$ For partial isometries: $$\tau^{(\prime)}:=J^*\sigma^{(\prime)}:\quad\sigma\perp\sigma'\implies ...


0

The assertion is false as: Given the Hilbert space $\ell^2(\mathbb{N})$. Consider the right shift: $$A:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad A:=R$$ Then it has finite trace: $$\sum_n\langle A\delta_n,\delta_n\rangle=\sum_n0=0$$ But it is not trace class: $$\sum_n\langle|A|\delta_n,\delta_n\rangle=\sum_n1=\infty$$ Concluding counterexample.


0

Meanwhile I got it... Limit Projection By orthogonality one has: $$\varphi\in\mathcal{R}^\perp\implies\varphi\in\mathcal{R}_\lambda^\perp\quad(\lambda\in\Lambda)$$ So one obtains: $$\|P_\lambda\varphi-0\|=\|0-0\|=0\stackrel{\lambda}{\to}0$$ By monotony one has: ...


1

The solution sets generally won't be the same. Try the example $$ A = \pmatrix{0&1\\1&0}, \\ h_1 = 0, \quad h_2 = 1 $$ In this case, the first equation gives us $$ \Re\{\psi_1\psi_2\} = 0 $$ Whereas the second gives us $$ \Im\{\psi_1\psi_2\} = 0 $$ The two equations have the same solution if and only if $A = 0$. Otherwise, suppose that $A_{ij} ...


1

Seems to me that the definition given for $H^s$ can't be right for $s<0$; it "must" be that $H^s$ is actually the space of tempered distributions $f$ such that $\hat f\in L^2(\mu)$, where $d\mu(\xi)=(1+|\xi|^2)^{s/2}\,d\xi$. Assuming so, this is easy: Say $(f_n)$ is Cauchy in $H^s$. Then $(\hat f_n)$ is Cauchy in $L^2(\mu)$. So $\hat f_n\to g$ in ...


1

The thing about $C(\Omega)$ works because $\mathbb R$ is complete: If $(f_n)$ is a Cauchy sequence in $C(\Omega)$ then the definition of the metric shows that for every $x\in\Omega$ the sequence $(f_n(x))$ is a Cauchy sequence of reals. Since $\mathbb R$ is complete there exists a real number $y$ such that $f_n(x)\to y$; we define $f(x)=y$ and proceed. The ...


3

You know the $u_m$ are converging to a function, because $\mathbb{R}$ is complete. For every $x \in \Omega$, you have $$| u_n(x) - u_m(x) | \leq \| u_n - u_m\|_\infty \to 0$$ So $u_n(x)$ is a cauchy sequence, and by completness of $\mathbb{R}$, converge. The function $u$ is then defined by : $$u(x) = \lim_{n\to \infty} u_n (x)$$ For your second ...


4

Let $x \in \mathcal{H}$ with $\lVert x\rVert = 1$. Show that $x$ cannot be in the orthogonal complement of $\operatorname{span} \{ f_n : n \in \mathbb{N}\}$. Since $\{ e_n : n \in \mathbb{N}\}$ is a Hilbert basis of $\mathcal{H}$, we can write $$x = \sum_{n = 0}^\infty c_n\cdot e_n.$$ Consider now $$y = \sum_{n = 0}^\infty c_n \cdot f_n.$$ Then we have ...


1

Consider the pushforward: $$E_\eta:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E_\eta(A):=E(\eta^{-1}A)$$ Their domain agree as:* $$\int|\vartheta\circ\eta|^2\mathrm{d}\nu_\varphi(\lambda)=\int|\vartheta|^2\mathrm{d}\nu^\eta_\varphi$$ Denote for shorthand: ...


1

In $M_2(\mathbb C)$, the normal operators are precisely the unitarily diagonalizable ones. So, any non-selfadjoint normal operator is, as TrialAndError mentioned, of the form $$ N=U^*\,\begin{bmatrix}\lambda&0\\0&\mu\end{bmatrix}\,U $$ with at least one of $\lambda,\mu$ not real (if both were real, $N=N^*$). The general $2\times 2$ unitary is of ...


1

In general, they are not weakly closed (not even strongly closed). To see this, consider the Hilbert space $\ell^2([0,1])$ and the operator $N=M_{\rm id}$. Then the (bounded) Borel functional calculus will yield all operators $M_f$ (multiplication with $f$, where $f$ is Borel measurable and bounded. But now consider the set $A:=\{I\subset [0,1]\,\mid\,I ...


2

This is a (probably icomplete) hierarchy of vector spaces from the point of view of functional analysis. Vector spaces: algebraic structure (addition and multiplication by scalars) Topological vector space: vector space with a topology such that addition and multiplication by scalars are continuous. Locally convex topological vector spaces: TVS in which ...


1

On $H=\mathbb C^2$, for each $\lambda\in \mathbb C$ let $P_\lambda$ be the projection onto the span of $\{(1,\lambda)\}$. Then $P_\lambda\to P_0$ as $\lambda\to 0$, but for $\lambda\neq 0$, $P_0\land P_\lambda =0$, while $P_0\land P_0=P_0\neq 0$. Also, for $\lambda\neq 0$, $P_0\lor P_\lambda = I$, while $P_0\lor P_0 = P_0\neq I$.


0

Meanwhile I got it... Define the function: $$\varphi\in\mathcal{D}H_0^2:\quad\omega(t):=U(t)^*JU_0(t)\varphi$$ Its derivatives are: $$\omega'(t)=U(t)i\{HJ-JH_0\}U_0(t)\varphi$$ $$\omega''(t)=U(t)i^2\{H^2J-2HJH_0+JH_0^2\}U_0(t)\varphi$$ By Pettis' criterion:* ...


0

Suppose a sequence $(h_n+Th_n)$ in $\hat{H}$ converges to some $x$, i.e., $h_n+Th_n\to x$. Applying $T$ and using the fact that $T^2=id$, we obtain $$Tx=\lim T(h_n+Th_n)=\lim Th_n+h_n=x,$$ so $x=\frac{x+Tx}{2}=\frac{x}{2}+T\frac{x}{2}\in\hat{H}$, which shows that $\hat{H}$ is closed.


1

The result is indeed true. We have $S = TT^* - T^*T$. As you said, we may conclude that $\langle x, Sx \rangle = 0$ for all $x \in X$, which in this case is a real vector space. We may use this information conclude that $S$ is skew-adjoint, that is, $S^* = -S$. We then additionally note that $$ S^* = (TT^*)^* - (T^*T)^* = TT^* - T^*T = S $$ So that $S$ ...


0

Meanwhile I got it... Projection They have domain: $$\mathcal{D}(P^*P)=\mathcal{D}(P^2)=\mathcal{D}P$$ So it must be bounded: $$\|P\varphi\|^2=\langle P^*P\varphi,\varphi\rangle=\langle P\varphi,\varphi\rangle\leq\|P\varphi\|\cdot\|\varphi\|$$ Also dense and closed: $$P=P^*=P^{**}\implies P=\overline{P}\quad(\overline{\mathcal{D}P}=\mathcal{H})$$ By ...


1

For the "only if" direction, consider the homogenous system $Ax = 0$. Then the $i$-th row is $$\sum_{j=1}^n \langle h_j, h_i \rangle x_j = 0$$ By linearity in the first argument it is $$\langle \sum_{j=1}^n x_j h_j, h_i \rangle = 0$$ Therefore $\sum_{j=1}^n x_j h_j \in \text{span}(h_1,h_2,\ldots,h_n)^\perp$. On the other hand $\sum_{j=1}^n x_j h_j$ is a ...


0

Note that one has: $$(\mathcal{N}\Omega)^\perp=\overline{\mathcal{R}\Omega^*}=\overline{\mathcal{R}|\Omega|}$$ Denote embeddings: ...


0

Denote for shorthand: $$\Omega':=\Omega^*\Omega\in\mathcal{B}(\mathcal{H}):\quad |\Omega|=\lim_n\Omega'_n$$ Regard spectral measures: $$H=\int\lambda\mathrm{d}E(\lambda)\quad H_0=\int\lambda\mathrm{d}E_0(\lambda)$$ By a previous thread:* $$\Omega E_0(A)=E(A)\Omega\implies\Omega'E_0(A)=E_0(A)\Omega'\implies(\Omega')^n E_0=E_0(A)(\Omega')^n$$ So one arrives ...


2

We must assume ${\mathcal D}A$ is dense, otherwise $A^*$ is not well defined. If $x \in {\mathcal D}A$ and $Ax = 0$ then of course $A^*A = 0$. For the converse, note that $y \in {\mathcal D} A^*$ with $A^* y = z$ iff for every $w \in {\mathcal D} A$, $\langle A w, y\rangle = \langle w, z\rangle$. Now $x \in {\mathcal N} A^* A$ iff $x \in {\mathcal D} A$ ...


1

On the one hand: $$\varphi\in\mathcal{N}A\implies A\varphi=0\in\mathcal{D}A^*\implies\varphi\in\mathcal{N}(A^*A)$$ On the other hand: $$\varphi\in\mathcal{N}(A^*A)\implies\|A\varphi\|^2=\langle A^*A\varphi,\varphi\rangle=0\implies\varphi\in\mathcal{N}A$$ Concluding the assertion.


0

Rudin restricts to normals as... Definition Given a Hilbert space $\mathcal{H}$. Consider selfadjoint operators: $$H:\mathcal{D}H\subseteq\mathcal{H}\to\mathcal{H}:\quad H=H^*$$ Define positivity by:* $$H\geq0:\iff\sigma(H)\geq0$$ *(They may be unbounded!) Positive Operators Given a Hilbert space $\mathcal{H}$. Consider positive operators: ...


3

If $\{e_n\}$ is an orthonormal basis for a Hilbert space $H$, then $$H = \left\{ \sum_{n=1}^\infty \alpha_n e_n : \sum_{n=1}^\infty |\alpha_n|^2 < \infty \right\}.$$ For $f=\sum \alpha_n e_n \in H$ we formally define the operation: $$Af = \sum_{n=1}^\infty \lambda_n \alpha_n e_n.$$ This operator is linear, and it is a well defined map from $H \to H$ ...


1

Meanwhile I got it: False! Given the Hilbert space $\ell^2(\mathbb{N})$. Consider right shifts: $$R_n:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N}):\quad R_n:=R^n$$ They are isometric: $$\|R^n\varphi\|=\|\varphi\|\quad(\varphi\in\ell^2)$$ But only weakly $$R_n\rightharpoonup0\quad R_n\not\to0$$ Concluding counterexample. See also: Weak vs. Strong


0

By unitarity one has: $$\|\Omega\varphi\|=\lim_{t\to\infty}\|JU_0(t)\varphi\|$$ It can be written as: $$\|JU_0(t)\varphi\|^2=\langle\{J^*J-1\}U_0(t)\varphi,U_0(t)\varphi\rangle+\|\varphi\|^2$$ But it is bounded: $$\|U_0(t)\varphi\|_{t\in\mathbb{R}}\equiv\|\varphi\|_{t\in\mathbb{R}}<\infty$$ So one obtains: ...


1

Without loss of generality we can assume that $z=0$ (otherwise apply the proof to the operator $A-z$ instead). The sequence $\{f_n\}$ is Weyl, i.e. $\|f_n\|=1$ and $\|Af_n\|\to 0$ when $n\to+\infty$. The vector $(f_n,Af_n)$ belongs to the graph $\Gamma(A)$. Since $A=\bar A_0$ we know that $\Gamma(A)=\bar\Gamma(A_0)$, so we can approximate $(f_n,Af_n)$ ...



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